Chapter 14 Molecular Model of Matter

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1 Chapter 14 Molecular Model of Matter GOALS When you have mastered the contents of this chapter, you will be able to achieve the following goals: Definitions Define each of the following terms, and use it in an operational definition: rms velocity surface tension adhesion capillarity cohesion osmosis Molecular Model Explain the properties of fluids using the molecular model. Problems Solve problems that relate the gas laws to the properties of the gas and that involve surface tension, capilarity, and osmosis. PREREQUISITES Before beginning this chapter you should have achieved the goals of Chapter 4, Forces and Newton's Laws, Chapter 5, Energy, Chapter 6, Momentum and Impulse, Chapter 10, Temperature and Heat, and Chapter 13, Elastic Properties of Materials 118

2 Chapter 14 Molecular Model of Matter OVERVIEW - Molecules are extremely small. Only the very largest molecules have been photographed and only with indirect methods using an electron microscope. The model scientists use to describe molecules and their movements and interactions is only an ideal representation. However, this model helps us gain an understanding for phenomena we can observe for a large number of molecules acting together. SUGGESTED STUDY PROCEDURE - Begin this chapter by reading the following Chapter Goals: Definitions, Molecular Model, and Problems. For additional discussion of the terms listed under the goal of Definitions, turn to the Definitions section of this Study Guide chapter. Now, continue by reading text sections In section 14.3 you may have difficulties following all the assumptions and mathematical steps. Even though this may be the case, pay close attention to the results as expressed by equations (14.14), (14.15), and (14.18). The example on page 312 will help clarify the use of the results derived in this section. At the end of the chapter, read the Chapter Summary and complete Summary Exercises Check your answers carefully against those provided. Next, do Algorithmic Problems 1-8 again checking your answers with those provided. Now do Exercises and Problems 1, 4, 5, 6, 9, 10, and 12. For additional work on the contents of this chapter, see the Examples section of this Study Guide chapter. Now you should be prepared to attempt the Practice Test on Molecular Model of Matter provided at the end of this Study Guide chapter. If you have difficulties with any of the concepts involved, please refer to the appropriate section of the text or this for more assistance. This study procedure is outlined below Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems Definitions 14.1,14.6, Molecular Model 14.2,14.3,14.4, Problems 14.8,14.9, ,4,5,6, 9,10,12 119

3 DEFINITIONS ROOT MEAN SQUARE VELOCITY - The square root of the sum of the squares of the velocities divided by the number of the particles. If we have two particles, one with a velocity of -20 m/s and the other with +20 m/s, their mean velocity is zero; their rms is +20 m/s. If we have two particles, one with velocity +50 m/s, the other with a velocity of +100 m/s; their mean velocity is +75m/s; their rms velocity is 56 m/s. ADHESION - Attraction of unlike molecules. The new epoxy cements have brought a whole era of adhesives for daily use. COHESION - Attraction of like molecules. Cohesive forces in solids are stronger than cohesive forces in liquids. SURFACE TENSION - Force necessary to break a unit length of surface. Have you ever seen a water spider "skate" along on the surface of a pond? CAPILLARITY - The rise of water in a capillary above the water level in the surrounding area. This arises when a liquid adheres to the walls of a narrow tube. OSMOSIS - Motion of fluid through a semipermeable membrane until the potential energy on the two sides of the membrane is the same. Osmosis is extremely common in biological systems. EXAMPLES KINETIC THEORY OF GASES 1. Assume that all the molecules in a gas are travelling in the x direction (1/2 of the molecules are headed in the plus x direction and 1/2 of the molecules are headed in the minus x direction). Assume that all the molecules have the same magnitude of velocity, v, and that n molecules strike a wall of area A in a time t. (a) From what volume do the molecules come that reach the area A of the wall in time t? (Sketch this volume on the figure below.) (b) Assume that all n molecules strike the wall head-on, are perfectly elastic, and rebound with a velocity v in the opposite (or - x) direction. What is the impulse imparted to the wall? (mass of each molecule = m) (c) What is the pressure on the wall? (d) In part (a) above, what is the number of molecules per unit volume that produce the bombardment of the wall? (e) In a closed container which has a volume V and contains a total of N molecules the number of molecules per unit volume is N/V. Make the correct substitution into your answer of part (c) and find an expression for the pressure of a gas in terms of N and V as well as m and v. 120

4 What Data Are Given? There are n molecules of mass m travelling with velocity v in the positive x direction that strike an area A of the wall perpendicular to the x-axis in a time of t. What Data Are Implied? The molecules are perfectly elastic and obey the laws of classical physics during the motion and collisions. What Physics Principles Are Involved? The fundamental laws of motion and collisions will be needed. What Equations Are to be Used? distance = velocity time impulse = change in momentum = force time pressure = force/area Algebraic Solution (a) In a time t, the molecules moving with a velocity v will travel a distance of vt. So the n molecules that hit the wall in time t had to be within a distance vt of the wall. (b) For each molecule p = -mv = (mv) = -2mv Impulse of wall on molecule = -2mv Impulse of a molecule on wall = +2mv Total impulse on the wall = n(2mv) = 2nmv (c) Pressure = force / area = (Impulse/time)/area = (2nmv)/(At) (d) Number per unit volume = number/volume = (rn)/(avt) = rn then pressure = 2rnmv 2 (1) (e) In a closed container only 1/6 of the molecules will have a positive x component of velocity since there are six equivalent orthogonal directions, +x, -x, +y, -y, +z, -z. So the r in Equation (1) is only 1/6 N/V; thus pressure = 2 (1/6 x N/V)mv 2 = 1/3 ((Nmv 2 ) / (v)) (2) 2. By the middle of the nineteenth century, the ideal gas law was written as PV = nrt (3) where the three variables are the pressure P (dynes/cm 2 ), the volume V(cm 3 ), and the temperature T ( ø K). The quantity n is the number of moles of gas in the sample (dimensionless) and R is a universal constant simply called the gas constant. The experimental value of the gas constant R is R = 8.31 x 10 7 ergs/ ø K where an erg = 1 dyne cm Later work by Avogadro led to a re-expression of the same law as PV = N k T (4) where the quantity N is the total number of molecules in the sample of gas and k is Boltzmann's constant. The experimental value for k is k = 1.38 x ergs/ ø K (a) Derive an expression for N in terms of k, n, and R. (b) Avogadro's number, N o, is the number of molecules in a mole (N/n). From the above data, determine the numerical value of N o. Thinking About This Problem The problem does not involve any profound knowledge of physics; rather straightforward manipulation of Equations (3) and (4) is all that is necessary. pv = nrt = NkT (a) so N = nr/k (b) Avogadro's number No = N/n = R/k = (8.31 x 10 7 ergs/ ø K)/(1.38 x ergs/ ø K) = 6.02 x

5 PRACTICE TEST 1. In an interesting physics demonstration, a young physics teacher placed a thin film of water between two flat panes of glass. The result is a bond which holds the two pieces of glass firmly together. Is this an example of adhesion or cohesion? Explain your answer as if you were teaching the class. 2. The rms speed of oxygen (O 2 ) molecules at 27 ø C is 482 m/sec. Assume that this is an ideal gas for the following questions. (The Boltzmann constant is 1.38 x J/molecule- ø k) a. Find the mass of the oxygen molecule in kilograms. b. Calculate the energy of an individual oxygen molecule. c. If the energy is to be reduced by a factor of 3, what reduction in temperature is required? 3. The surface tension of an unknown liquid is measured at room temperature with the apparatus shown below. First the system is balanced using a small counter balance m. After the ring of radius r (10 cm) is placed in the unknown liquid, an additional mass of Δm must be added to pull the circular wire from the liquid. If m = 50.2 grams and Δm = 18.8 grams, calculate the surface tension of the liquid at room temperature. (Express your answer in dynes/cm.) ANSWERS: 1. Adhesion. Water molecules adhere (are attracted) to glass molecules. The forces between individual water molecules are cohesive forces x kg, 6.2 x Joule, Reduce temperature by 200 ø K dynes/cm 122

Revision Guide for Chapter 13

Revision Guide for Chapter 13 Matter: very simple Revision Guide for Chapter Contents Student s Checklist Revision Notes Ideal gas... Ideal gas laws... Assumptions of kinetic theory of gases... 5 Internal energy... 6 Specific thermal